# I think my teacher is wrong, do you agree with me???

#### ironsheep

##### Junior Member
For the question,

She says the answer is C because

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#### MarkFL

##### Super Moderator
Staff member
I agree with your teacher that C is the correct choice. Why do you disagree?

#### ironsheep

##### Junior Member
I think it is 6 to the power of (5/6) because of

#### MarkFL

##### Super Moderator
Staff member
If you're going to do it like that (which is perfectly fine), then consider:

$$\displaystyle \sqrt[3]{6\sqrt{6}}=\sqrt[3]{6}\cdot\sqrt[3]{\sqrt{6}}=\left(6^{\frac{1}{3}}\right)\left(\left(6^{\frac{1}{2}}\right)\right)^{\frac{1}{3}}=6^{\frac{1}{3}}\cdot6^{\frac{1}{6}}=6^{\frac{1}{3}+\frac{1}{6}}=6^{\frac{2}{6}+\frac{1}{6}}=6^{\frac{3}{6}}=6^{\frac{1}{2}}=\sqrt{6}$$

#### Jomo

##### Elite Member
But that sqrt(6) is under the cube root sign.

#### ironsheep

##### Junior Member
I am still confused. My method seems flawless and her method seems overlycomplicated.

#### Jomo

##### Elite Member
Please re-read my post above. What you are missing is that everything under the cube root sign needs to be cube rooted.

You need to use the fact that cuberoot (A*B) = cuberoot (A)* cuberoot(B)

You have cuberoot[6*sqrt(6)] = cuberoot(6)*cuberoot(sqrt(6)) = 61/3* cuberoot(61/2) = 61/3* (61/2)1/3 = 61/3*61/6 = 61/3 + 1/6 = 61/2 = sqrt(6)

#### ironsheep

##### Junior Member
No, then it would be

#### pka

##### Elite Member
I am still confused. My method seems flawless and her method seems overlycomplicated.
Why in the world would you think that your sequence is flawless?

\displaystyle \begin{align*}\large\sqrt[3]{{6\sqrt 6 }}&={\left[ {6{{\left( 6 \right)}^{\frac{1}{2}}}} \right]^{\frac{1}{3}}}\\&={\left[ {{6^{\frac{3}{2}}}} \right]^{\frac{1}{3}}}\\&=\left[ {{6^{\frac{1}{2}}}} \right]\\&=\sqrt 6 \end{align*}

#### ironsheep

##### Junior Member
I still don't understand. I just don't get it.

#### JeffM

##### Elite Member

{{6)(1/2)}^(1/3) = 6^(1/6).

And then as others have pointed out

6^(1/3) * 6^(1/6) = 6^(2/6) * 6^(1/6) = 6^{(2+1)/6} = 6^(3/6) = 6^(1/2).

#### ironsheep

##### Junior Member
I still don't understand---- maybe if you explained it in detail better or something. Is anyone here now how to explain things well because I don't understand still and these posts aren't helping.

#### Harry_the_cat

##### Senior Member
Ironsheep, people are really trying to help you here, and it has been explained in detail. Don't take out your frustration on those trying to help!
Here's another way (similar to some above):
$$\displaystyle \sqrt[3]{6*\sqrt{6}} = \sqrt[3]{6^1*6^{\frac{1}{2}}}$$ …… (1)

$$\displaystyle =\sqrt[3]{6^{\frac{3}{2}}}$$ ….. (2)

$$\displaystyle =(6^{\frac{3}{2}})^{\frac{1}{3}}$$ ….. (3)

$$\displaystyle =6^{\frac{3}{2} * \frac{1}{3}}$$ ….. (4)

$$\displaystyle = 6^{\frac{1}{2}}$$ …. (5)

$$\displaystyle = \sqrt{6}$$ ….. (6)

I've labelled each step with numbers. Instead of saying "I don't get it" tell us which step you don't understand.

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#### ironsheep

##### Junior Member
Thank you. I don't get step one, why are you putting a one on 6 and why you putting both numbers under the same line??

#### JeffM

##### Elite Member
The radical is a grouping indicator like parentheses.

$$\displaystyle \sqrt[3]{xy}$$ calls for taking the cube root of BOTH x and y.

That is why they are under the same line. It is just like the line that separates the numerator and denominator of a fraction.

$$\displaystyle 5 = \dfrac{12 + 8}{3 + 1} \ne 12 = 4 + 8 = \dfrac{12}{3}+\dfrac{8}{1}.$$

The line in both cases says you must consider the expression as a whole.

And do you not know that $$\displaystyle a = a^1?$$

#### ironsheep

##### Junior Member
I did not know that 6= 6 to the power of one. Also, why do you add like bases and not multiply them?? Also, for step one, why can't you separate them out like I did in my images??

#### JeffM

##### Elite Member
You need to learn the rules of exponents.

Assuming $$\displaystyle a,\ b,\ \text {and } c \text { are all integers, and } a > 0, \text { then}$$

$$\displaystyle a^0 = 1;$$

$$\displaystyle \text {Integer } n > 0 \implies a * a^{(n-1)} \implies$$

$$\displaystyle a^1 = a * 1 = a,\ a^2 = a * a^1 = a * a,\ a^3 = a * a^2 = a * a * a,\ \text {etc;}$$

$$\displaystyle a^b * a^c = a^{(b+c)};$$

$$\displaystyle a^{-c} = \dfrac{1}{a^c} = \left ( \dfrac{1}{a} \right )^c;$$

$$\displaystyle \dfrac{a^b}{a^c} = a^{(b-c)};$$

$$\displaystyle (a^b)^c = a^{(bc)};$$

$$\displaystyle a^{(1/c)} = \sqrt[c]{a}; \text { and}$$

$$\displaystyle a^{(b/c)} = \sqrt[c]{a^b}.$$

#### ironsheep

##### Junior Member
Why are the rules the way they are?? What determines them to be true, like if you gave someone two pieces of candy and then gave them another two pieces of candy, then you gave that person 4 pieces of candy( not 5, not 4.5, not 4.1). So when would all these exponents be actually used in real life?? What is the practical day to day approach to this stuff?

Also, for step one, why can't you separate them out like I did in my images?? Why are the numbers trapped under the hood sign thing??

#### JeffM

##### Elite Member
Why are the rules the way they are?? What determines them to be true, like if you gave someone two pieces of candy and then gave them another two pieces of candy, then you gave that person 4 pieces of candy( not 5, not 4.5, not 4.1). So when would all these exponents be actually used in real life?? What is the practical day to day approach to this stuff?

Also, for step one, why can't you separate them out like I did in my images?? Why are the numbers trapped under the hood sign thing??
I explained to you that the radical is in part a grouping symbol.

What does $$\displaystyle \sqrt[3]{6}$$ mean?

It means a number such that $$\displaystyle \sqrt[3]{6} * \sqrt[3]{6} * \sqrt[3]{6} = 6.$$

What does $$\displaystyle \sqrt{6}$$ mean?

It means a number such that $$\displaystyle \sqrt{6} * \sqrt{6} = 6.$$

Is it not obvious then that $$\displaystyle 6^{(1/3)} = \sqrt[3]{6} \ne \sqrt{6} = 6^{(1/2)}?$$

They are two different numbers. For a clearer example,

$$\displaystyle \sqrt[3]{64} = 4 \text { because } 4 * 4 * 4 = 64.$$

$$\displaystyle \sqrt{64} = 8 \text { because } 8 * 8 = 64.$$

$$\displaystyle 4 \ne 8.$$

Basically, the rules of exponents are a bunch of definitions. They fit logically together, but they are not discoveries, but creations. Exponents are a convenient notation to express certain kinds of multiplicative relations.

#### ironsheep

##### Junior Member
Why do people need to learn this stuff? Why the struggle over something I just need to learn to pass tests and pass school and then forget about as if it doesn't exist?

Was anything in my images remotely correct?